An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-6.3,0)\) and \((6.3,0)\) and a covertex at \((0, 6)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.7, 5.05):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(6.3)^2 + (6)^2 =a^2\]
\[a ~=~ \sqrt{6.3^2+6^2} ~=~ 8.7\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,6) to \(F_2\)\[2a = 17.4\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=6.3\]\[2c = 12.6\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.7-(-6.3))^2+5.05^2}\]
\[s=12.1038217\]\[s\approx12.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.7-(6.3))^2+5.05^2}\]
\[w=5.297405\]\[w\approx5.3\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=17.4\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4,0)\) and \((4,0)\) and a covertex at \((0, 4.2)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.21, 2.89):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4)^2 + (4.2)^2 =a^2\]
\[a ~=~ \sqrt{4^2+4.2^2} ~=~ 5.8\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,4.2) to \(F_2\)\[2a = 11.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4\]\[2c = 8\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.21-(-4))^2+2.89^2}\]
\[s=8.7038038\]\[s\approx8.7\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.21-(4))^2+2.89^2}\]
\[w=2.8976197\]\[w\approx2.9\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=11.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-3.6,0)\) and \((3.6,0)\) and a covertex at \((0, 7.7)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (7.31, 3.93):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(3.6)^2 + (7.7)^2 =a^2\]
\[a ~=~ \sqrt{3.6^2+7.7^2} ~=~ 8.5\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,7.7) to \(F_2\)\[2a = 17\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=3.6\]\[2c = 7.2\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(7.31-(-3.6))^2+3.93^2}\]
\[s=11.5962494\]\[s\approx11.6\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(7.31-(3.6))^2+3.93^2}\]
\[w=5.4045351\]\[w\approx5.4\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=17\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-1.1,0)\) and \((1.1,0)\) and a covertex at \((0, 6)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (5.53, 2.53):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(1.1)^2 + (6)^2 =a^2\]
\[a ~=~ \sqrt{1.1^2+6^2} ~=~ 6.1\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,6) to \(F_2\)\[2a = 12.2\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=1.1\]\[2c = 2.2\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(5.53-(-1.1))^2+2.53^2}\]
\[s=7.096323\]\[s\approx7.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(5.53-(1.1))^2+2.53^2}\]
\[w=5.1015488\]\[w\approx5.1\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=12.2\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4.2,0)\) and \((4.2,0)\) and a covertex at \((0, 4)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (3.17, 3.35):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4.2)^2 + (4)^2 =a^2\]
\[a ~=~ \sqrt{4.2^2+4^2} ~=~ 5.8\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,4) to \(F_2\)\[2a = 11.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4.2\]\[2c = 8.4\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(3.17-(-4.2))^2+3.35^2}\]
\[s=8.0956408\]\[s\approx8.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(3.17-(4.2))^2+3.35^2}\]
\[w=3.5047682\]\[w\approx3.5\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=11.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-2.8,0)\) and \((2.8,0)\) and a covertex at \((0, 4.5)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (3.6, 3.3):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(2.8)^2 + (4.5)^2 =a^2\]
\[a ~=~ \sqrt{2.8^2+4.5^2} ~=~ 5.3\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,4.5) to \(F_2\)\[2a = 10.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=2.8\]\[2c = 5.6\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(3.6-(-2.8))^2+3.3^2}\]
\[s=7.2006944\]\[s\approx7.2\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(3.6-(2.8))^2+3.3^2}\]
\[w=3.3955854\]\[w\approx3.4\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=10.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-2.1,0)\) and \((2.1,0)\) and a covertex at \((0, 7.2)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.3, 5.9):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(2.1)^2 + (7.2)^2 =a^2\]
\[a ~=~ \sqrt{2.1^2+7.2^2} ~=~ 7.5\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,7.2) to \(F_2\)\[2a = 15\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=2.1\]\[2c = 4.2\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.3-(-2.1))^2+5.9^2}\]
\[s=8.7045965\]\[s\approx8.7\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.3-(2.1))^2+5.9^2}\]
\[w=6.2968246\]\[w\approx6.3\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=15\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4.5,0)\) and \((4.5,0)\) and a covertex at \((0, 2.8)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.59, 1.4):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4.5)^2 + (2.8)^2 =a^2\]
\[a ~=~ \sqrt{4.5^2+2.8^2} ~=~ 5.3\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,2.8) to \(F_2\)\[2a = 10.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4.5\]\[2c = 9\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.59-(-4.5))^2+1.4^2}\]
\[s=9.1971789\]\[s\approx9.2\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.59-(4.5))^2+1.4^2}\]
\[w=1.4028899\]\[w\approx1.4\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=10.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-6.5,0)\) and \((6.5,0)\) and a covertex at \((0, 7.2)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (8.96, 2.76):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(6.5)^2 + (7.2)^2 =a^2\]
\[a ~=~ \sqrt{6.5^2+7.2^2} ~=~ 9.7\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,7.2) to \(F_2\)\[2a = 19.4\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=6.5\]\[2c = 13\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(8.96-(-6.5))^2+2.76^2}\]
\[s=15.7044325\]\[s\approx15.7\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(8.96-(6.5))^2+2.76^2}\]
\[w=3.6971881\]\[w\approx3.7\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=19.4\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-2.8,0)\) and \((2.8,0)\) and a covertex at \((0, 9.6)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.99, 8.32):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(2.8)^2 + (9.6)^2 =a^2\]
\[a ~=~ \sqrt{2.8^2+9.6^2} ~=~ 10\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,9.6) to \(F_2\)\[2a = 20\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=2.8\]\[2c = 5.6\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.99-(-2.8))^2+8.32^2}\]
\[s=11.3976533\]\[s\approx11.4\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.99-(2.8))^2+8.32^2}\]
\[w=8.6034005\]\[w\approx8.6\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=20\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
So:
\[\begin{align}
h &= -2\\\\
k &= -5\\\\
r_1 &= 7\\\\
r_2 &= 3
\end{align}\]
Substitute those numbers into the equation:
\[\frac{(x+2)^2}{49}+\frac{(y+5)^2}{9}=1\]
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 16 meters apart, and she uses a string that is 20 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=20\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 4.8 meters apart, and she uses a string that is 14.8 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=14.8\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 11.2 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 3.2 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 2.6 meters apart, and she uses a string that is 17 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 14.4 meters apart, and she uses a string that is 18 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=18\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 2.6 meters apart, and she uses a string that is 17 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 13 meters apart, and she uses a string that is 19.4 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=19.4\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 6.6 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 11 meters apart, and she uses a string that is 14.6 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=14.6\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 9.6 meters apart, and she uses a string that is 14.6 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 14.6 meters long; divide by 2 to find \(a\).
\[a = \frac{14.6}{2} = 7.3\]
The stakes are 9.6 meters apart; divide by 2 to find \(c\).
\[c = \frac{9.6}{2} = 4.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[4.8^2+b^2=7.3^2 \]
\[b^2=7.3^2-4.8^2\]
\[b^2=53.29-23.04\]
\[b^2=30.25\]
\[b=\sqrt{30.25}\]
\[b=5.5\]
The length of the minor axis is twice \(b\).
\[2b=11\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 3.2 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The stakes are 3.2 meters apart; divide by 2 to find \(c\).
\[c = \frac{3.2}{2} = 1.6\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[1.6^2+b^2=6.5^2 \]
\[b^2=6.5^2-1.6^2\]
\[b^2=42.25-2.56\]
\[b^2=39.69\]
\[b=\sqrt{39.69}\]
\[b=6.3\]
The length of the minor axis is twice \(b\).
\[2b=12.6\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 2.2 meters apart, and she uses a string that is 12.2 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 12.2 meters long; divide by 2 to find \(a\).
\[a = \frac{12.2}{2} = 6.1\]
The stakes are 2.2 meters apart; divide by 2 to find \(c\).
\[c = \frac{2.2}{2} = 1.1\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[1.1^2+b^2=6.1^2 \]
\[b^2=6.1^2-1.1^2\]
\[b^2=37.21-1.21\]
\[b^2=36\]
\[b=\sqrt{36}\]
\[b=6\]
The length of the minor axis is twice \(b\).
\[2b=12\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 13 meters apart, and she uses a string that is 19.4 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 19.4 meters long; divide by 2 to find \(a\).
\[a = \frac{19.4}{2} = 9.7\]
The stakes are 13 meters apart; divide by 2 to find \(c\).
\[c = \frac{13}{2} = 6.5\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.5^2+b^2=9.7^2 \]
\[b^2=9.7^2-6.5^2\]
\[b^2=94.09-42.25\]
\[b^2=51.84\]
\[b=\sqrt{51.84}\]
\[b=7.2\]
The length of the minor axis is twice \(b\).
\[2b=14.4\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 10.8 meters apart, and she uses a string that is 18 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 18 meters long; divide by 2 to find \(a\).
\[a = \frac{18}{2} = 9\]
The stakes are 10.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{10.8}{2} = 5.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[5.4^2+b^2=9^2 \]
\[b^2=9^2-5.4^2\]
\[b^2=81-29.16\]
\[b^2=51.84\]
\[b=\sqrt{51.84}\]
\[b=7.2\]
The length of the minor axis is twice \(b\).
\[2b=14.4\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 12.8 meters apart, and she uses a string that is 16 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 16 meters long; divide by 2 to find \(a\).
\[a = \frac{16}{2} = 8\]
The stakes are 12.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{12.8}{2} = 6.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.4^2+b^2=8^2 \]
\[b^2=8^2-6.4^2\]
\[b^2=64-40.96\]
\[b^2=23.04\]
\[b=\sqrt{23.04}\]
\[b=4.8\]
The length of the minor axis is twice \(b\).
\[2b=9.6\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 5.6 meters apart, and she uses a string that is 10.6 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 10.6 meters long; divide by 2 to find \(a\).
\[a = \frac{10.6}{2} = 5.3\]
The stakes are 5.6 meters apart; divide by 2 to find \(c\).
\[c = \frac{5.6}{2} = 2.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[2.8^2+b^2=5.3^2 \]
\[b^2=5.3^2-2.8^2\]
\[b^2=28.09-7.84\]
\[b^2=20.25\]
\[b=\sqrt{20.25}\]
\[b=4.5\]
The length of the minor axis is twice \(b\).
\[2b=9\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 7.8 meters apart, and she uses a string that is 17.8 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 17.8 meters long; divide by 2 to find \(a\).
\[a = \frac{17.8}{2} = 8.9\]
The stakes are 7.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{7.8}{2} = 3.9\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[3.9^2+b^2=8.9^2 \]
\[b^2=8.9^2-3.9^2\]
\[b^2=79.21-15.21\]
\[b^2=64\]
\[b=\sqrt{64}\]
\[b=8\]
The length of the minor axis is twice \(b\).
\[2b=16\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 9.6 meters apart, and she uses a string that is 14.6 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 14.6 meters long; divide by 2 to find \(a\).
\[a = \frac{14.6}{2} = 7.3\]
The stakes are 9.6 meters apart; divide by 2 to find \(c\).
\[c = \frac{9.6}{2} = 4.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[4.8^2+b^2=7.3^2 \]
\[b^2=7.3^2-4.8^2\]
\[b^2=53.29-23.04\]
\[b^2=30.25\]
\[b=\sqrt{30.25}\]
\[b=5.5\]
The length of the minor axis is twice \(b\).
\[2b=11\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 12.8 meters apart, and she uses a string that is 16 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 16 meters long; divide by 2 to find \(a\).
\[a = \frac{16}{2} = 8\]
The stakes are 12.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{12.8}{2} = 6.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.4^2+b^2=8^2 \]
\[b^2=8^2-6.4^2\]
\[b^2=64-40.96\]
\[b^2=23.04\]
\[b=\sqrt{23.04}\]
\[b=4.8\]
The length of the minor axis is twice \(b\).
\[2b=9.6\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 16.8 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 9 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=10.6\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 14.6 meters and the length of the minor axis to be 9.6 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=14.6\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 19.4 meters and the length of the minor axis to be 14.4 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=19.4\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 5.6 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=10.6\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17.8 meters and the length of the minor axis to be 16 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17.8\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 15.4 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 15.4 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 12.6 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 15 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17.8 meters and the length of the minor axis to be 7.8 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 17.8 meters long, then the string is 17.8 meters long; divide by 2 to find \(a\).
\[a = \frac{17.8}{2} = 8.9\]
The length minor axis is 7.8 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{7.8}{2} = 3.9\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[3.9^2+c^2=8.9^2 \]
\[c^2=8.9^2-3.9^2\]
\[c^2=79.21-15.21\]
\[c^2=64\]
\[c=\sqrt{64}\]
\[c=8\]
The distance between the stakes is \(2c\).
\[2c=16\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 19.4 meters and the length of the minor axis to be 13 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 19.4 meters long, then the string is 19.4 meters long; divide by 2 to find \(a\).
\[a = \frac{19.4}{2} = 9.7\]
The length minor axis is 13 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{13}{2} = 6.5\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.5^2+c^2=9.7^2 \]
\[c^2=9.7^2-6.5^2\]
\[c^2=94.09-42.25\]
\[c^2=51.84\]
\[c=\sqrt{51.84}\]
\[c=7.2\]
The distance between the stakes is \(2c\).
\[2c=14.4\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 9 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 10.6 meters long, then the string is 10.6 meters long; divide by 2 to find \(a\).
\[a = \frac{10.6}{2} = 5.3\]
The length minor axis is 9 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{9}{2} = 4.5\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[4.5^2+c^2=5.3^2 \]
\[c^2=5.3^2-4.5^2\]
\[c^2=28.09-20.25\]
\[c^2=7.84\]
\[c=\sqrt{7.84}\]
\[c=2.8\]
The distance between the stakes is \(2c\).
\[2c=5.6\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 11.2 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 13 meters long, then the string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The length minor axis is 11.2 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{11.2}{2} = 5.6\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[5.6^2+c^2=6.5^2 \]
\[c^2=6.5^2-5.6^2\]
\[c^2=42.25-31.36\]
\[c^2=10.89\]
\[c=\sqrt{10.89}\]
\[c=3.3\]
The distance between the stakes is \(2c\).
\[2c=6.6\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 5.6 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 10.6 meters long, then the string is 10.6 meters long; divide by 2 to find \(a\).
\[a = \frac{10.6}{2} = 5.3\]
The length minor axis is 5.6 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{5.6}{2} = 2.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[2.8^2+c^2=5.3^2 \]
\[c^2=5.3^2-2.8^2\]
\[c^2=28.09-7.84\]
\[c^2=20.25\]
\[c=\sqrt{20.25}\]
\[c=4.5\]
The distance between the stakes is \(2c\).
\[2c=9\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 6.6 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 13 meters long, then the string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The length minor axis is 6.6 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{6.6}{2} = 3.3\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[3.3^2+c^2=6.5^2 \]
\[c^2=6.5^2-3.3^2\]
\[c^2=42.25-10.89\]
\[c^2=31.36\]
\[c=\sqrt{31.36}\]
\[c=5.6\]
The distance between the stakes is \(2c\).
\[2c=11.2\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 12.6 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 13 meters long, then the string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The length minor axis is 12.6 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{12.6}{2} = 6.3\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.3^2+c^2=6.5^2 \]
\[c^2=6.5^2-6.3^2\]
\[c^2=42.25-39.69\]
\[c^2=2.56\]
\[c=\sqrt{2.56}\]
\[c=1.6\]
The distance between the stakes is \(2c\).
\[2c=3.2\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17.8 meters and the length of the minor axis to be 16 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 17.8 meters long, then the string is 17.8 meters long; divide by 2 to find \(a\).
\[a = \frac{17.8}{2} = 8.9\]
The length minor axis is 16 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{16}{2} = 8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[8^2+c^2=8.9^2 \]
\[c^2=8.9^2-8^2\]
\[c^2=79.21-64\]
\[c^2=15.21\]
\[c=\sqrt{15.21}\]
\[c=3.9\]
The distance between the stakes is \(2c\).
\[2c=7.8\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 16.8 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 17 meters long, then the string is 17 meters long; divide by 2 to find \(a\).
\[a = \frac{17}{2} = 8.5\]
The length minor axis is 16.8 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{16.8}{2} = 8.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[8.4^2+c^2=8.5^2 \]
\[c^2=8.5^2-8.4^2\]
\[c^2=72.25-70.56\]
\[c^2=1.69\]
\[c=\sqrt{1.69}\]
\[c=1.3\]
The distance between the stakes is \(2c\).
\[2c=2.6\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 19 meters and the length of the minor axis to be 15.2 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 19 meters long, then the string is 19 meters long; divide by 2 to find \(a\).
\[a = \frac{19}{2} = 9.5\]
The length minor axis is 15.2 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{15.2}{2} = 7.6\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[7.6^2+c^2=9.5^2 \]
\[c^2=9.5^2-7.6^2\]
\[c^2=90.25-57.76\]
\[c^2=32.49\]
\[c=\sqrt{32.49}\]
\[c=5.7\]
The distance between the stakes is \(2c\).
\[2c=11.4\]
I should also mention that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[36x^2+72x+4y^2-24y-72=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[36(x^2+2x)+4(y^2-6y)=72\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((x+1)^2\equiv x^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 36, we are technically adding 36 to both sides to make an equivalent equation.
\[36(x^2+2x+1)+4(y^2-6y)=72+36\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-6\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (9). Notice \((y-3)^2\equiv y^2-6x+9\). Lastly, since we are adding 9 inside parentheses with a multiplier of 4, we are technically adding 36 to both sides to make an equivalent equation.
\[36(x^2+2x+1)+4(y^2-6y+9)=72+36+36\]
Combine the constants.
\[36(x^2+2x+1)+4(y^2-6y+9)=144\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[36(x+1)^2+4(y-3)^2=144\]
Divide both sides by \(144\).
\[\frac{36(x+1)^2}{144}+\frac{4(y-3)^2}{144}=1\]
Simplify the fractions. Notice that \(\frac{36}{144}= \frac{1}{4}\) and \(\frac{4}{144}= \frac{1}{36}\).
\[\frac{(x+1)^2}{4}+\frac{(y-3)^2}{36}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{4}=2\) and \(\sqrt{36}=6\).
\[\frac{(x+1)^2}{2^2}+\frac{(y-3)^2}{6^2}=1\]
Thus,
\[\begin{align}
h &= -1 \\
k &= 3 \\
r_1 &= 2 \\
r_2 &= 6
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[49x^2+196x+25y^2-50y-1004=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[49x^2+196x+25y^2-50y-1004=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[49(x^2+4x)+25(y^2-2y)=1004\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(4\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (4). Notice \((x+2)^2\equiv x^2+4x+4\). Lastly, since we are adding 4 inside parentheses with a multiplier of 49, we are technically adding 196 to both sides to make an equivalent equation.
\[49(x^2+4x+4)+25(y^2-2y)=1004+196\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((y-1)^2\equiv y^2-2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 25, we are technically adding 25 to both sides to make an equivalent equation.
\[49(x^2+4x+4)+25(y^2-2y+1)=1004+196+25\]
Combine the constants.
\[49(x^2+4x+4)+25(y^2-2y+1)=1225\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[16x^2-32x+36y^2-144y-416=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[16(x^2-2x)+36(y^2-4y)=416\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((x-1)^2\equiv x^2-2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 16, we are technically adding 16 to both sides to make an equivalent equation.
\[16(x^2-2x+1)+36(y^2-4y)=416+16\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-4\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (4). Notice \((y-2)^2\equiv y^2-4x+4\). Lastly, since we are adding 4 inside parentheses with a multiplier of 36, we are technically adding 144 to both sides to make an equivalent equation.
\[16(x^2-2x+1)+36(y^2-4y+4)=416+16+144\]
Combine the constants.
\[16(x^2-2x+1)+36(y^2-4y+4)=576\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[16(x-1)^2+36(y-2)^2=576\]
Divide both sides by \(576\).
\[\frac{16(x-1)^2}{576}+\frac{36(y-2)^2}{576}=1\]
Simplify the fractions. Notice that \(\frac{16}{576}= \frac{1}{36}\) and \(\frac{36}{576}= \frac{1}{16}\).
\[\frac{(x-1)^2}{36}+\frac{(y-2)^2}{16}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{36}=6\) and \(\sqrt{16}=4\).
\[\frac{(x-1)^2}{6^2}+\frac{(y-2)^2}{4^2}=1\]
Thus,
\[\begin{align}
h &= 1 \\
k &= 2 \\
r_1 &= 6 \\
r_2 &= 4
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[25x^2-150x+36y^2+72y-639=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[25x^2-150x+36y^2+72y-639=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[25(x^2-6x)+36(y^2+2y)=639\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-6\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (9). Notice \((x-3)^2\equiv x^2-6x+9\). Lastly, since we are adding 9 inside parentheses with a multiplier of 25, we are technically adding 225 to both sides to make an equivalent equation.
\[25(x^2-6x+9)+36(y^2+2y)=639+225\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((y+1)^2\equiv y^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 36, we are technically adding 36 to both sides to make an equivalent equation.
\[25(x^2-6x+9)+36(y^2+2y+1)=639+225+36\]
Combine the constants.
\[25(x^2-6x+9)+36(y^2+2y+1)=900\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[25(x-3)^2+36(y+1)^2=900\]
Divide both sides by \(900\).
\[\frac{25(x-3)^2}{900}+\frac{36(y+1)^2}{900}=1\]
Simplify the fractions. Notice that \(\frac{25}{900}= \frac{1}{36}\) and \(\frac{36}{900}= \frac{1}{25}\).
\[\frac{(x-3)^2}{36}+\frac{(y+1)^2}{25}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{36}=6\) and \(\sqrt{25}=5\).
\[\frac{(x-3)^2}{6^2}+\frac{(y+1)^2}{5^2}=1\]
Thus,
\[\begin{align}
h &= 3 \\
k &= -1 \\
r_1 &= 6 \\
r_2 &= 5
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[4x^2+24x+16y^2-256y+996=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[4x^2+24x+16y^2-256y+996=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[4(x^2+6x)+16(y^2-16y)=-996\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(6\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (9). Notice \((x+3)^2\equiv x^2+6x+9\). Lastly, since we are adding 9 inside parentheses with a multiplier of 4, we are technically adding 36 to both sides to make an equivalent equation.
\[4(x^2+6x+9)+16(y^2-16y)=-996+36\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-16\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (64). Notice \((y-8)^2\equiv y^2-16x+64\). Lastly, since we are adding 64 inside parentheses with a multiplier of 16, we are technically adding 1024 to both sides to make an equivalent equation.
\[4(x^2+6x+9)+16(y^2-16y+64)=-996+36+1024\]
Combine the constants.
\[4(x^2+6x+9)+16(y^2-16y+64)=64\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[4(x+3)^2+16(y-8)^2=64\]
Divide both sides by \(64\).
\[\frac{4(x+3)^2}{64}+\frac{16(y-8)^2}{64}=1\]
Simplify the fractions. Notice that \(\frac{4}{64}= \frac{1}{16}\) and \(\frac{16}{64}= \frac{1}{4}\).
\[\frac{(x+3)^2}{16}+\frac{(y-8)^2}{4}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{16}=4\) and \(\sqrt{4}=2\).
\[\frac{(x+3)^2}{4^2}+\frac{(y-8)^2}{2^2}=1\]
Thus,
\[\begin{align}
h &= -3 \\
k &= 8 \\
r_1 &= 4 \\
r_2 &= 2
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[9x^2+18x+81y^2-648y+576=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[9x^2+18x+81y^2-648y+576=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[9(x^2+2x)+81(y^2-8y)=-576\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((x+1)^2\equiv x^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 9, we are technically adding 9 to both sides to make an equivalent equation.
\[9(x^2+2x+1)+81(y^2-8y)=-576+9\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-8\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (16). Notice \((y-4)^2\equiv y^2-8x+16\). Lastly, since we are adding 16 inside parentheses with a multiplier of 81, we are technically adding 1296 to both sides to make an equivalent equation.
\[9(x^2+2x+1)+81(y^2-8y+16)=-576+9+1296\]
Combine the constants.
\[9(x^2+2x+1)+81(y^2-8y+16)=729\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[9(x+1)^2+81(y-4)^2=729\]
Divide both sides by \(729\).
\[\frac{9(x+1)^2}{729}+\frac{81(y-4)^2}{729}=1\]
Simplify the fractions. Notice that \(\frac{9}{729}= \frac{1}{81}\) and \(\frac{81}{729}= \frac{1}{9}\).
\[\frac{(x+1)^2}{81}+\frac{(y-4)^2}{9}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{81}=9\) and \(\sqrt{9}=3\).
\[\frac{(x+1)^2}{9^2}+\frac{(y-4)^2}{3^2}=1\]
Thus,
\[\begin{align}
h &= -1 \\
k &= 4 \\
r_1 &= 9 \\
r_2 &= 3
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[9x^2+18x+36y^2+504y+1449=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[9x^2+18x+36y^2+504y+1449=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[9(x^2+2x)+36(y^2+14y)=-1449\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((x+1)^2\equiv x^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 9, we are technically adding 9 to both sides to make an equivalent equation.
\[9(x^2+2x+1)+36(y^2+14y)=-1449+9\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(14\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (49). Notice \((y+7)^2\equiv y^2+14x+49\). Lastly, since we are adding 49 inside parentheses with a multiplier of 36, we are technically adding 1764 to both sides to make an equivalent equation.
\[9(x^2+2x+1)+36(y^2+14y+49)=-1449+9+1764\]
Combine the constants.
\[9(x^2+2x+1)+36(y^2+14y+49)=324\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[9(x+1)^2+36(y+7)^2=324\]
Divide both sides by \(324\).
\[\frac{9(x+1)^2}{324}+\frac{36(y+7)^2}{324}=1\]
Simplify the fractions. Notice that \(\frac{9}{324}= \frac{1}{36}\) and \(\frac{36}{324}= \frac{1}{9}\).
\[\frac{(x+1)^2}{36}+\frac{(y+7)^2}{9}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{36}=6\) and \(\sqrt{9}=3\).
\[\frac{(x+1)^2}{6^2}+\frac{(y+7)^2}{3^2}=1\]
Thus,
\[\begin{align}
h &= -1 \\
k &= -7 \\
r_1 &= 6 \\
r_2 &= 3
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[16x^2-32x+36y^2-360y+340=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[16x^2-32x+36y^2-360y+340=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[16(x^2-2x)+36(y^2-10y)=-340\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((x-1)^2\equiv x^2-2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 16, we are technically adding 16 to both sides to make an equivalent equation.
\[16(x^2-2x+1)+36(y^2-10y)=-340+16\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-10\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (25). Notice \((y-5)^2\equiv y^2-10x+25\). Lastly, since we are adding 25 inside parentheses with a multiplier of 36, we are technically adding 900 to both sides to make an equivalent equation.
\[16(x^2-2x+1)+36(y^2-10y+25)=-340+16+900\]
Combine the constants.
\[16(x^2-2x+1)+36(y^2-10y+25)=576\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[16(x-1)^2+36(y-5)^2=576\]
Divide both sides by \(576\).
\[\frac{16(x-1)^2}{576}+\frac{36(y-5)^2}{576}=1\]
Simplify the fractions. Notice that \(\frac{16}{576}= \frac{1}{36}\) and \(\frac{36}{576}= \frac{1}{16}\).
\[\frac{(x-1)^2}{36}+\frac{(y-5)^2}{16}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{36}=6\) and \(\sqrt{16}=4\).
\[\frac{(x-1)^2}{6^2}+\frac{(y-5)^2}{4^2}=1\]
Thus,
\[\begin{align}
h &= 1 \\
k &= 5 \\
r_1 &= 6 \\
r_2 &= 4
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[36x^2-288x+9y^2+18y+261=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[36x^2-288x+9y^2+18y+261=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[36(x^2-8x)+9(y^2+2y)=-261\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(-8\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (16). Notice \((x-4)^2\equiv x^2-8x+16\). Lastly, since we are adding 16 inside parentheses with a multiplier of 36, we are technically adding 576 to both sides to make an equivalent equation.
\[36(x^2-8x+16)+9(y^2+2y)=-261+576\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((y+1)^2\equiv y^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 9, we are technically adding 9 to both sides to make an equivalent equation.
\[36(x^2-8x+16)+9(y^2+2y+1)=-261+576+9\]
Combine the constants.
\[36(x^2-8x+16)+9(y^2+2y+1)=324\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[36(x-4)^2+9(y+1)^2=324\]
Divide both sides by \(324\).
\[\frac{36(x-4)^2}{324}+\frac{9(y+1)^2}{324}=1\]
Simplify the fractions. Notice that \(\frac{36}{324}= \frac{1}{9}\) and \(\frac{9}{324}= \frac{1}{36}\).
\[\frac{(x-4)^2}{9}+\frac{(y+1)^2}{36}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{9}=3\) and \(\sqrt{36}=6\).
\[\frac{(x-4)^2}{3^2}+\frac{(y+1)^2}{6^2}=1\]
Thus,
\[\begin{align}
h &= 4 \\
k &= -1 \\
r_1 &= 3 \\
r_2 &= 6
\end{align}\]
Question
The following equation (in polynomial form) represents an ellipse.
\[9x^2+72x+36y^2+72y-144=0\]
With some algebra (completing the square), you can convert the equation into standard form:
From the given polynomial form, determine the 4 parameters.
\(h=\)
\(k=\)
\(r_1=\)
\(r_2=\)
Solution
\[9x^2+72x+36y^2+72y-144=0\]
Factor out the quadratic coefficients from each variable’s polynomial. Also move the constant term to the right with addition/subtration.
\[9(x^2+8x)+36(y^2+2y)=144\]
Complete the square for the \(x\) quadratic. If \(x^2+bx+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(8\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (16). Notice \((x+4)^2\equiv x^2+8x+16\). Lastly, since we are adding 16 inside parentheses with a multiplier of 9, we are technically adding 144 to both sides to make an equivalent equation.
\[9(x^2+8x+16)+36(y^2+2y)=144+144\]
Complete the square for the \(y\) quadratic. If \(y^2+by+c\) is a perfect square, then \(c=\left(\frac{b}{2}\right)^2\). In this case, the linear coefficient is \(2\). If you divide it by 2 and square the result, you get the necessary constant to complete the square (1). Notice \((y+1)^2\equiv y^2+2x+1\). Lastly, since we are adding 1 inside parentheses with a multiplier of 36, we are technically adding 36 to both sides to make an equivalent equation.
\[9(x^2+8x+16)+36(y^2+2y+1)=144+144+36\]
Combine the constants.
\[9(x^2+8x+16)+36(y^2+2y+1)=324\]
O yeah, remember how you completed the square? Twice? Let’s rewrite those quadratic expressions in factored form. (This could have been done earlier.)
\[9(x+4)^2+36(y+1)^2=324\]
Divide both sides by \(324\).
\[\frac{9(x+4)^2}{324}+\frac{36(y+1)^2}{324}=1\]
Simplify the fractions. Notice that \(\frac{9}{324}= \frac{1}{36}\) and \(\frac{36}{324}= \frac{1}{9}\).
\[\frac{(x+4)^2}{36}+\frac{(y+1)^2}{9}=1\]
If you find the square roots of both denominators, you will have the standard form of an ellipse. Notice \(\sqrt{36}=6\) and \(\sqrt{9}=3\).
\[\frac{(x+4)^2}{6^2}+\frac{(y+1)^2}{3^2}=1\]
Thus,
\[\begin{align}
h &= -4 \\
k &= -1 \\
r_1 &= 6 \\
r_2 &= 3
\end{align}\]
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(11.6\) meters and two stakes placed \(8\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{11.6}{2}~=~5.8\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{8}{2}=4\]
We still need \(b\), the shorter radius.
\[b=4.2\]
Find the area.
\[A ~=~ ab\pi ~=~ (5.8)(4.2)\pi\]
\[A \approx 76.53\]
The area is approximately 76.53 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(14.6\) meters and two stakes placed \(11\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{14.6}{2}~=~7.3\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{11}{2}=5.5\]
We still need \(b\), the shorter radius.
\[b=4.8\]
Find the area.
\[A ~=~ ab\pi ~=~ (7.3)(4.8)\pi\]
\[A \approx 110.08\]
The area is approximately 110.08 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(19.4\) meters and two stakes placed \(14.4\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{19.4}{2}~=~9.7\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{14.4}{2}=7.2\]
We still need \(b\), the shorter radius.
\[b=6.5\]
Find the area.
\[A ~=~ ab\pi ~=~ (9.7)(6.5)\pi\]
\[A \approx 198.08\]
The area is approximately 198.08 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(13.6\) meters and two stakes placed \(6.4\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{13.6}{2}~=~6.8\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{6.4}{2}=3.2\]
We still need \(b\), the shorter radius.
\[b=6\]
Find the area.
\[A ~=~ ab\pi ~=~ (6.8)(6)\pi\]
\[A \approx 128.18\]
The area is approximately 128.18 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(10.6\) meters and two stakes placed \(9\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{10.6}{2}~=~5.3\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{9}{2}=4.5\]
We still need \(b\), the shorter radius.
\[b=2.8\]
Find the area.
\[A ~=~ ab\pi ~=~ (5.3)(2.8)\pi\]
\[A \approx 46.62\]
The area is approximately 46.62 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(17.4\) meters and two stakes placed \(12\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{17.4}{2}~=~8.7\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{12}{2}=6\]
We still need \(b\), the shorter radius.
\[b=6.3\]
Find the area.
\[A ~=~ ab\pi ~=~ (8.7)(6.3)\pi\]
\[A \approx 172.19\]
The area is approximately 172.19 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(14.6\) meters and two stakes placed \(11\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{14.6}{2}~=~7.3\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{11}{2}=5.5\]
We still need \(b\), the shorter radius.
\[b=4.8\]
Find the area.
\[A ~=~ ab\pi ~=~ (7.3)(4.8)\pi\]
\[A \approx 110.08\]
The area is approximately 110.08 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(17.8\) meters and two stakes placed \(7.8\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{17.8}{2}~=~8.9\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{7.8}{2}=3.9\]
We still need \(b\), the shorter radius.
\[b=8\]
Find the area.
\[A ~=~ ab\pi ~=~ (8.9)(8)\pi\]
\[A \approx 223.68\]
The area is approximately 223.68 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(10.4\) meters and two stakes placed \(4\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{10.4}{2}~=~5.2\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
\[c=\frac{4}{2}=2\]
We still need \(b\), the shorter radius.
\[b=4.8\]
Find the area.
\[A ~=~ ab\pi ~=~ (5.2)(4.8)\pi\]
\[A \approx 78.41\]
The area is approximately 78.41 square meters.
Question
It is easy to find the area of an ellipse. It is surprisingly difficult to find the perimeter (you need calculus).
The area of an ellipse is pretty intuitive because an ellipse is a stretched circle. The ellipse has two different radii, so instead of \(A=\pi\cdot r\cdot r\) (for a circle), an ellipse’s area is found with the following equation.
\[A=\pi a b\]
where \(a\) is the length of the semi-major axis and \(b\) is the length of the semi-minor axis. Or, \(a\) and \(b\) are the largest and smallest radii of the ellipse.
If an ellipse is drawn with a string of length \(10.6\) meters and two stakes placed \(9\) meters apart. What is the area of the ellipse in square meters?
(The tolerance is \(\pm 0.01 ~\mathrm{m}^2\).)
Solution
We need to find \(a\) and \(b\). Remember the string’s length equals \(2a\).
\[a~=~\frac{2a}{2}~=~\frac{10.6}{2}~=~5.3\]
The distance from the center to a focus, \(c\), equals half the distance between the stakes.
